Hey guys I want to learn how to calculate current values for a BJT in the linear region and to calculate the resistor values i require in order to run the transister at it's max collector current/voltage. Basically for the bc639, 640 and bd140 and 139 transistors. any help would be greatly appreciated. cheers.

 

You can't operate a transistor at both its max current and its max voltage. It will get too hot and will fail.
You must calculate its max power dissipation with a suitable heatsink, then calculate a current and a voltage for it.

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Uncle $crooge

 

so for the max power rating which rating do we use on the data sheet? Is it the Ptot at Tc < 25 degrees or Tamb < 25 degrees? and when you say heat sink do u mean a resistor or something else? and once i find out the voltages and currents I am able to pass through how do i go about calculating the base and emitter resistor values?

 

A transistor's power rating is with the temperature of its case (Tc) somehow kept at only 25 degrees C. A heatsink is a finned aluminum extrusion that can carry away a certain amount of heat from a transistor that is bolted to it, but isn't perfect so the case of the transistor heats above 25 degrees C which reduces its max power rating.

A 115W rated 2N3055 transistor will be near its max temperature if it dissipates only about 70W on a huge heatsink and without an insulator. A fan will help the heatsink get rid of the heat.

You need to learn about how transistors work, their specifications and about Ohm's Law for you to calculate resistor values to bias a transistor.

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Uncle $crooge

 

ok well i just calculated that my transistor won't be running at full spec most of the time so it shouldn't need a heat sink? but when the motor (the load I am driving) stalls it draws a very high current and pretty much puts the transister at it's max power rating. but this is only for a few seconds, so do i still need a heat sink for this situation even though it is only for a few seconds? and will the exposure to this high power every now and then damage my transistor eventually? Also are the calculations for a transistor in the linear region the same as for saturation with the exception of the 0.7V drop across the transistor? cheers.

 

You have to ask yourself one question. Will the transistor be more likely to survive with or without a heatsink? Do you feel lucky today...?

 

Yes, power transistors are designed to be mounted on heatsinks for VERY good reason - they generally dissipate very little power without one.

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Yet again another vague request. What transistor are you using at what current and power level?

If youre planning to control the speed of a DC motor you should bias the transistor as a switch, not in the linear region and use PWM to control the speed. Also, a MOSFET is probably far more suited to this application.

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Yes they are different as shown on the transistor's datasheet.
It takes a lot more base current to make a transistor saturate so it is turned on hard, than to operate it at the same current in its linear region.

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Uncle $crooge

 

I am using a high power transistor (BD139 and BD140) the Ptot at Tc < 25 is 12.5W and my supply varies between 14-24V depending on situations and the max stall current of my motor is 600mA. I am also using PWM to drive the motor. The reason why I am using the linear region is because it was easier to implement the H-bridge to drive the motor using a conventional push-pull set up (I had problems implementing a collector-collector H-bridge). I know this means I will lose 1.4V across the transistor but I think my system might be able to deal with this loss. If I use my transistors in this way will I be pushing them too hard? and possibly reduce their lifetime? cheers.

 

Honestly dialect beside me doubts on count that that you has written,
since I found on sites 1
, 2 in which write about quite other...

 

Wthout seeing your schematic attached here we are just guessing. You are using PWM but the transistors are linear? Then are they emitter-followers?

What pulls up the base of the NPN? What pulls down the base of the PNP?

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Uncle $crooge

 

i've attached a schematic of the configuration i'm using. Yea, i'm using PWM and running them in the linear region, dunno if this is a good idea or not. but the drive circuit is much simpler and seems to work fine. problem is i'm not too sure on the calculations on this circuit?

Attached Images

h-bridge.JPG (20.6 KB, 26 views)

 

You have a 1k resistor providing the 60mA to the base of the BD139, therefore the loss in the 1k resistor is 1k x 60mA= 60V! See, an emitter follower has too high a voltage drop unless its base resistor is from a very high voltage.
If a common emitter PNP transistor is used with its collector pulling up the 600mA motor's wire then its 60mA base current can come through a resistor from ground and it will turn on properly.

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Uncle $crooge

 

o whoops, i forgot to change the values. I am currently using 270 ohms instead of 1k and 1.8k instead of 10k. my input signals to the drive circuit are also 4.2V. So this shud mean the voltage drop is 270 x 60mA = 16.2V right? and i'm also guessing your assuming the B gain of the transister is 100? thats how u got 60mA right? also by the sounds of it this is totally the wrong way to run this motor as i'm pushing everything too hard. by calculation this circuit should be burning out? unless i'm doing the calculations all wrong? (o btw it doesn't always run at 600mA it only hits that value for a few seconds before the system kills the signal)